NCERT Solutions for Class 11 Maths Chapter 5-Complex Numbers and Quadratic Equations Exercise 5.3

The National Council of Educational Research and Training was established by the government with the intention of improving the quality of Indian education. CBSE is responsible for governing the NCERT. There is a Central Board of Secondary Education in India known as CBSE. A CBSE board is governed by the government of India. It is mandatory for students to achieve a minimum score of 33% on both the theory and practical examinations during Class 11 in order to be promoted to Class 12. A fundamental part of the human intellect and logic is Mathematics, which underpins our understanding of the world. In order to simplify Complex Numbers and Quadratic Equations Class 11 Exercise 5.3 it is often necessary to utilise mathematical concepts. Understanding mathematical concepts and terminology is essential for students who wish to pursue careers in Science, Engineering, Commerce, Scientist, Astronomy, and Economics, among other fields.

In previous classes, the students learned about linear equations in one and two variables, along with the concept of quadratic equations.  Complex numbers and quadratic equations is a branch of mathematics that deals with important theorems, concepts, and formulae.It includes linear and quadratic equations, as well as roots related to the complex number set called complex roots. Students may be able to solve all the questions presented in the exercises in this chapter at a steady pace if they use the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3. Students can also understand the chapters’ real-world applications as a result of the concepts discussed in NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3. The NCERT Solutions can also be used to prepare for a variety of competitive examinations, such as JEE, BITSAT, and other college entrance exams.

The NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 provided by Extramarks are curated by skilled specialists to assist students in understanding concepts in a more efficient and effective manner. There is an in-depth, step-by-step explanation of textbook problems in NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3. The Extramarks study materials provide revision notes, sample problems, past papers, and worksheets related to Class 11 Mathematics for students preparing for their final exams.

Besides providing NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3, Extramarks also provides NCERT Solutions Class 11 which contains answers to all the chapters and exercises present in the Class 11 Mathematics Textbook. Besides providing solutions for Class 11, students from other classes can also access the NCERT solutions prepared by Extramarks.

All students from primary, secondary and high schools are provided with the advantage to refer to these solutions prepared by subject experts. NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4, NCERT Solutions Class 5, NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions Class 9, NCERT Solutions Class 10, NCERT Solutions Class 12.

The students are encouraged to make use of the Extramarks’ NCERT solutions in order to prepare for their exams in all subjects. There is a link on Extramarks’ website that allows them to download these solutions. Moreover, students consider these solutions to be reliable information resources since they have been prepared by experts in the field.

The concepts and questions from the NCERT textbooks for Class 11 Mathematics have been considered throughout NCERT Solutions for Class 11 Maths, Chapter 5 Exercise 5.3.A PDF of NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 can also be downloaded from the Extramarks website. There may be an advantage to using Extramarks’ study materials for students who have difficulty comprehending the concepts. In accordance with the revised guidelines and syllabus for Class 11 Mathematics, NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 have been prepared by subject experts. These solutions allow students to practise answering questions and gain a deeper understanding of the subject.

NCERT Solutions for Class 11 Maths Chapter 5-Complex Numbers and Quadratic Equations Exercise 5.3

To prepare for the Class 11 examinations, Extramarks recommends downloading the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3, which is available in PDF format. In the event that students do not always have access to the internet, they can use the downloaded PDF of NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 whenever and wherever they are located. Extramarks has prepared these NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 with the assistance of subject experts with extensive knowledge of the subject. These solutions can be downloaded by students and accessed offline as well. The NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 should be reviewed by students as they prepare for the Class 11 Mathematics final exam.

They can download the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 by clicking here:

Access NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations.

NCERT Solutions for Class 11 Maths Chapters

The NCERT Solutions for Class 11 Mathematics are available on Extramarks for all chapters in the NCERT textbook. Following is a list of the solutions for each chapter of the book:

NCERT Solution Class 11 Maths of Chapter 5 Exercise

There is a table below that containsquestions from all the exercises included in Class 11 Mathematics Chapter 5 that can be useful to students who are having difficulty finding the solutions and questions.

Class 11 Maths NCERT Solutions Chapter 5 Exercise 5.3

Class 11 is an important year for a student. It is the decision-making class in which they determine which subject stream they intend to pursue for their higher studies. The importance of Class 11 cannot be overstated, as it determines the future careers of the students. Even though it is not the board year, it helps them prepare for it.Having a solid understanding of the concepts taught in Class 11 allows them to learn more complex subjects in Class 12.

Although many students utilize the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 when studying for Chapter 5 of Class 11 Mathematics, they are underappreciated.

NCERT textbooks are among the most commonly used study materials in Class 11, which is why we are seeing a high demand for NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3. It is essential to review the Maths NCERT Solutions Class 11 Chapter 5 Exercise 5.3 in order to prepare for the Class 11 final exams. Students may find Extramarks’ NCERT Solutions For Class 11 Mathematics and other learning resources helpful in preparing for their Class 11 examinations. A simple check of the answer alone does not provide sufficient information regarding how the solution was reached.

Students may find it helpful to refer to NCERT Solutions for a better understanding of the procedure and steps. This enables students to answer questions on the final examination with greater ease. Through the use of NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3, students can gain a comprehensive understanding of the chapter. A better understanding of linear inequalities can be gained by reviewing each solution.

Students are advised to take assistance from the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 before answering the questions. Further, students should be able to comprehend the principles presented in this exercise. It is generally beneficial to solve exercises from the NCERT book because these problems are frequently asked in exams. It should be noted, however, that some of the questions in this exercise may be difficult for students to answer. Thus, Extramarks provides NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 for their convenience.

CBSE Maths Class 11 NCERT Solutions Chapter 5 Exercise 5.3

The NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 include all potential questions from the Class 11 Mathematics NCERT textbook. Preparing for exams with these solutions helps the students  get acquainted with the CBSE question pattern and the types of questions that are likely to appear in a CBSE examination. Class 11 solutions provide comprehensive coverage of all concepts in various chapters to ensure that students do not miss any information.

Extramarks’ Class 11 Mathematics NCERT Solutions Chapter 5.3 has been prepared in a step-by-step format to facilitate understanding of the exercise. They provide many study resources for thorough exam preparation. The subject experts ensure that Extramarks’ NCERT Solutions and study materials are conceptually clear and explained well in simple language. The NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 include all solved examples as well as step-by-step solutions to all exercise questions.

An Overview of the Important Topics Covered in Exercise 5.3 of Class 11 Maths NCERT Solutions

Extramarks provides students with NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 through their website and mobile application. It is safe for students to consider the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3  a reliable source of information due to the fact that they are curated by subject experts. Detailed explanations of NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 are provided in accordance with NCERT textbooks. It is possible for students to find NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 on the Extramarks website. There are questions included in this exercise regarding finding the complex roots of quadratic equations. Additionally, it contains numerous examples and problems that illustrate how complex numbers can be used in quadratic equations. A deeper understanding of the concepts can be gained by students on a regular practice of NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3.

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The Extramarks learning platform provides students with access to curriculum-mapped learning modules. Moreover, they provide NCERT Solutions for all classes and subjects for the purpose of assisting students with their homework and exam preparation. Various study materials are available on the online platform to assist students in preparing for the exam. The following are examples of study materials: past years’ papers, sample papers, assignments, worksheets, and doubt-solving groups. Through Extramarks’ Learning App, students are able to bridge the gap between their strong areas and areas in need of improvement. As a result, they are able to respond confidently during the examination for Class 11.

Q.1 Solve the following equation:
x2 + 3 = 0

Ans.

x 2 +3 =0 comparing with ax 2 +bx+c=0, we get a=1,b=0 and c=3 By using quadratic formula,       x= −b± b 2 −4ac 2a = −0± 0 2 −4( 1 )( 3 ) 2( 1 ) =± −12 2 =± 2i 3 2 [ ∵ −1 =i ]      x=±i 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@CACA@

Q.2 Solve the following equation:
2x2 + x + 1 = 0

Ans.

2x 2 +x+1=0 comparing with ax 2 +bx+c=0, we get a=2,  b=1 and c=1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6DD7@ By using quadratic formula,       x= −b± b 2 −4ac 2a = −1± 1 2 −4( 2 )( 1 ) 2( 2 ) = −1± 1−8 4 = −1± −7 4      x= −1±i 7 4 [ ∵ −1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9F8C@

Q.3 Solve the following equation:
x2 + 3x + 9 = 0

Ans.

x 2 +3x+9=0 comparing with ax 2 +bx+c=0, we get a=1,  b=3 and c=9 By using quadratic formula,       x= −b± b 2 −4ac 2a = −3± 3 2 −4( 1 )( 9 ) 2( 1 ) = −3± 9−36 2 = −3± −27 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@BDCB@      x= −3±3 3  i 2 [ ∵ −1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadIhacqGH9aqpdaWcaaqaaiabgkHiTiaaiodacqGHXcqScaaIZaWaaOaaaeaacaaIZaaaleqaaOGaaGPaVlaadMgaaeaacaaIYaaaaiaaxMaacaWLjaWaamWaaeaacqWI1isudaGcaaqaaiabgkHiTiaaigdaaSqabaGccqGH9aqpcaWGPbaacaGLBbGaayzxaaaaaa@5372@

Q.4 Solve the following equation:
– x2 + x – 2 = 0

Ans.

−  x 2 +x−2=0 comparing with ax 2 +bx+c=0, we get a=−1,  b=1 and c=−2 By using quadratic formula,       x= −b± b 2 −4ac 2a = −1± 1 2 −4( −1 )( −2 ) 2( −1 ) = −1± 1−8 −2 = −1± −7 −2      x= −1±i 7 −2 [ ∵ −1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DD93@

Q.5 Solve the following equation:
x2 + 3x + 5 = 0

Ans.

x 2 +3x+5=0 comparing with ax 2 +bx+c=0, we get a=1,  b=3 and c=5 By using quadratic formula, MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@85D8@       x= −b± b 2 −4ac 2a = −3± 3 2 −4( 1 )( 5 ) 2( 1 ) = −3± 9−20 2 = −1± −11 2      x= −3±i 11 2 [ ∵ −1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaabIhacqGH9aqpdaWcaaqaaiabgkHiTiaadkgacqGHXcqSdaGcaaqaaiaadkgadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamyyaiaadogaaSqabaaakeaacaaIYaGaamyyaaaaaeaacaWLjaGaeyypa0ZaaSaaaeaacqGHsislcaaIZaGaeyySae7aaOaaaeaacaaIZaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGinamaabmaabaGaaGymaaGaayjkaiaawMcaamaabmaabaGaaGynaaGaayjkaiaawMcaaaWcbeaaaOqaaiaaikdadaqadaqaaiaaigdaaiaawIcacaGLPaaaaaaabaGaaCzcaiabg2da9maalaaabaGaeyOeI0IaaG4maiabgglaXoaakaaabaGaaGyoaiabgkHiTiaaikdacaaIWaaaleqaaaGcbaGaaGOmaaaacqGH9aqpdaWcaaqaaiabgkHiTiaaigdacqGHXcqSdaGcaaqaaiabgkHiTiaaigdacaaIXaaaleqaaaGcbaGaaGOmaaaaaeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadIhacqGH9aqpdaWcaaqaaiabgkHiTiaaiodacqGHXcqScaWGPbWaaOaaaeaacaaIXaGaaGymaaWcbeaaaOqaaiaaikdaaaGaaCzcaiaaxMaacaWLjaWaamWaaeaacqWI1isudaGcaaqaaiabgkHiTiaaigdaaSqabaGccqGH9aqpcaWGPbaacaGLBbGaayzxaaaaaaa@8A60@

Q.6 Solve the following equation:
x2 – x + 2 = 0

Ans.

x 2 –x+2=0 Comparing with ax 2 +bx+c=0, we get a=1,  b=−1 and c=2 By using quadratic formula,       x= −b± b 2 −4ac 2a = −( −1 )± ( −1 ) 2 −4( 1 )( 2 ) 2( 1 ) = 1± 1−8 2 = 1± −7 2      x= 1± 7  i 2 [ ∵ −1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D92C@

Q.7 Solve the following equation:

2 x 2 +x+ 2 =0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaGcaaqaaGqabiaa=jdaaSqabaGccaWF4bWaaWbaaSqabeaacGaMa+Nmaaaakiaa=TcacaWF4bGaa83kamaakaaabaGaa8NmaaWcbeaakiaa=1dacaWFWaaaaa@4189@

Ans.

2 x 2 +x+ 2 =0 Comparing with ax 2 +bx+c=0, we get a= 2 ,  b=1 and c= 2 By using quadratic formula,       x= −b± b 2 −4ac 2a = −1± 1 2 −4( 2 )( 2 ) 2( 2 ) = −1± 1−8 2 2 = −1± −7 2 2      x= −1±i 7 2 2 [ ∵ −1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaamaakaaabaGaaGOmaaWcbeaakiaadIhadaahaaWcbeqaaiacyciIYaaaaOGaey4kaSIaamiEaiabgUcaRmaakaaabaGaaGOmaaWcbeaakiabg2da9iaaicdaaeaacaWGdbGaam4Baiaad2gacaWGWbGaamyyaiaadkhacaWGPbGaamOBaiaadEgacaqGGaGaae4DaiaabMgacaqG0bGaaeiAaiaabccacaqGHbGaaeiEamaaCaaaleqabaGaaeOmaaaakiabgUcaRiaadkgacaWG4bGaey4kaSIaam4yaiabg2da9iaaicdacaGGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaabggacqGH9aqpdaGcaaqaaiaaikdaaSqabaGccaGGSaGaaGPaVlaaykW7caWGIbGaeyypa0JaaGymaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaae4yaiabg2da9maakaaabaGaaGOmaaWcbeaaaOqaaiaadkeacaWG5bGaaeiiaiaabwhacaqGZbGaaeyAaiaab6gacaqGNbGaaeiiaiaabghacaqG1bGaaeyyaiaabsgacaqGYbGaaeyyaiaabshacaqGPbGaae4yaiaabccacaqGMbGaae4BaiaabkhacaqGTbGaaeyDaiaabYgacaqGHbGaaeilaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaabIhacqGH9aqpdaWcaaqaaiabgkHiTiaadkgacqGHXcqSdaGcaaqaaiaadkgadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamyyaiaadogaaSqabaaakeaacaaIYaGaamyyaaaaaeaacaWLjaGaeyypa0ZaaSaaaeaacqGHsislcaaIXaGaeyySae7aaOaaaeaacaaIXaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGinamaabmaabaWaaOaaaeaacaaIYaaaleqaaaGccaGLOaGaayzkaaWaaeWaaeaadaGcaaqaaiaaikdaaSqabaaakiaawIcacaGLPaaaaSqabaaakeaacaaIYaWaaeWaaeaadaGcaaqaaiaaikdaaSqabaaakiaawIcacaGLPaaaaaaabaGaaCzcaiabg2da9maalaaabaGaeyOeI0IaaGymaiabgglaXoaakaaabaGaaGymaiabgkHiTiaaiIdaaSqabaaakeaacaaIYaWaaOaaaeaacaaIYaaaleqaaaaakiabg2da9maalaaabaGaeyOeI0IaaGymaiabgglaXoaakaaabaGaeyOeI0IaaG4naaWcbeaaaOqaaiaaikdadaGcaaqaaiaaikdaaSqabaaaaaGcbaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG4bGaeyypa0ZaaSaaaeaacqGHsislcaaIXaGaeyySaeRaamyAamaakaaabaGaaG4naaWcbeaaaOqaaiaaikdadaGcaaqaaiaaikdaaSqabaaaaOGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaWaamWaaeaacqWI1isudaGcaaqaaiabgkHiTiaaigdaaSqabaGccqGH9aqpcaWGPbaacaGLBbGaayzxaaaaaaa@DB95@

Q.8 Solve the following equation:

3 x 2 − 2 x+3 3 =0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaGcaaqaaGqabiaa=ndaaSqabaGccaWF4bWaaWbaaSqabeaacGaMa+NmaaaakiabgkHiTmaakaaabaGaa8NmaaWcbeaakiaa=HhacaWFRaGaa83mamaakaaabaGaa83maaWcbeaakiaa=1dacaWFWaaaaa@4358@

Ans.

3 x 2 − 2 x+3 3 =0 Comparing with ax 2 +bx+c=0, we get a= 3 ,  b=− 2 and c=3 3 By using quadratic formula,       x= −b± b 2 −4ac 2a MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A12A@ = −( − 2 )± ( − 2 ) 2 −4( 3 )( 3 3 ) 2( 3 ) = 2 ± 2−36 2 3 = 2 ± −34 2 3      x= 2 ± 34  i 2 3 [ ∵ −1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7C74@

Q.9 Solve the following equation:

x 2 +x+ 1 2 =0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaaieqacaWF4bWaaWbaaSqabeaacGaMa+Nmaaaakiaa=TcacaWF4bGaa83kamaalaaabaGaaGymaaqaamaakaaabaGaa8NmaaWcbeaaaaGccaWF9aGaa8hmaaaa@417C@

Ans.

x 2 +x+ 1 2 =0 Comparing with ax 2 +bx+c=0, we get a=1,  b=1 and c= 1 2 By using quadratic formula,       x= −b± b 2 −4ac 2a = −( 1 )± ( 1 ) 2 −4( 1 )( 1 2 ) 2( 1 ) = −1± 1−2 2 2 = −1± 1−2 2 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C4B1@      x= −1±i 2 2 −1 2 [ ∵ −1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadIhacqGH9aqpdaWcaaqaaiabgkHiTiaaigdacqGHXcqScaWGPbWaaOaaaeaacaaIYaWaaOaaaeaacaaIYaaaleqaaOGaeyOeI0IaaGymaaWcbeaaaOqaaiaaikdaaaGaaCzcaiaaxMaadaWadaqaaiablwJirnaakaaabaGaeyOeI0IaaGymaaWcbeaakiabg2da9iaadMgaaiaawUfacaGLDbaaaaa@53B0@

Q.10 Solve the following equation:

x2+x2+1=0

Ans.

x2+x2+1=0Comparing with ax2+bx+c=0, we geta=1,  b=12 and c=1By using quadratic formula,x=−b±b2−4ac2a =−(12)±(12)2−4(1)(1)2(1) =−12±12−42 =−12±−722     x=−1±i722 [∵ −1=i]

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FAQs (Frequently Asked Questions)

1. Is it possible for NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 to assist students in clarifying the concepts in Class 11 Mathematics?

The NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 can be used to practice questions after students have finished their preparation. It is intended to assist students in identifying and improving their mistakes, thus clarifying their concepts, using the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3. Additionally, Extramarks provides students with study materials and other resources to assist them in preparing for their exams.

2. How can students get access to NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 on the Internet?

There is an online version of the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 available at Extramarks for students in Class 11. The answers to textbook questions serve as a framework for understanding the topics presented in each chapter. Students may also use the questions as a source of learning material. On the Extramarks website, experts prepare NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3 to assist students in completing their curriculum in a timely manner. This allows students to focus on learning concepts more effectively, thus establishing a stronger foundation.

3. Among the other NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3, what makes Extramarks' NCERT solutions unique?

Among the most comprehensive and well-designed study guides available online for Class 11 5.3 Maths, Extramarks offers exercise-wise solutions. The Maths NCERT Solutions Class 11 Chapter 5 Exercise 5.3 provides step-by-step instructions for each problem. Using Extramarks’ NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3, students can easily revise, understand, and solve doubts. The NCERT Solutions Class 11 Maths Chapter 5 Exercise 5.3 are advantageous, as they are easy to access, can be downloaded offline and are prepared by subject experts.